   Chapter 28, Problem 65PE

Chapter
Section
Textbook Problem

A Van de Graaff accelerator utilizes a 50.0 MV potential difference to accelerate charged particles such as protons. (a) What is the velocity of a proton accelerated by such a potential? (b) An electron?

To determine

(a)

The velocity of proton.

Explanation

Given data:

Charge is q=1.60×1019C.

The velocity of light is c=3.00×108m/s.

Potential difference is V=50.0MV.

Formula used:

The expression for γis,

γ=11v2c2v=c11γ2

The relativistic kinetic energy is,

KE=(γ1)mc2    ....... (1)

The kinetic energy gained by particle of charge qis,

KE=qV    ...... (2)

Equate the equation (1) and (2).

qV=(γ1)mc2γ=qVmc2+1

Substitute the given value

To determine

(b)

The velocity of electron.

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